Question

Stomach acid, a dilute solution of $HCl$ in water, can be neutralized by reaction with sodium hydrogen carbonate, $NaHC{O}_{3(aq)}+HC{L}_{(aq)}⟶NaC{L}_{(aq)}+H_2{O}_{(l)}+C{O}_{2(g)}$
How many milliliters of $0.125MNaHC{O}_3$ solution are needed to neutralize $18.0mL$ of $0.100MHCL$ ?

• A. 14.4 mL
• B. 12.0 mL
• C. 14.0 mL
• D. 13.2 mL

Answer 1

Answer: (A)

Given, $M_{HCl}=0.1M,V_{HCl}=18.0mL$
$M_{NaHC{O}_3}=0.125M,V_{NaHC{O}_3}=?$
On applying, $M_{HCl}\times V_{HCl}=M_{NaHC{O}_3}\times V_{NaHC{O}_3}$
$\Rightarrow 0.1\times 18=0.125\times V_{NaHC{O}_3}$
$\Rightarrow M_{NaHC{O}_3}=14.4mL$
Thus, $14.4mL$ of the $1.25MNaHCO{}_3$ solution is needed to neutralise $18.0mL$ of the $0.100MHCl$ solution.